Question: A curve is defined by the parametric equations $x=3t-6$ and $y=e^{2t}+10$. What is $\dfrac{d^2y}{dx^2}$ in terms of $t$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{2e^{-2t}}{9}$ (Choice B) B $\dfrac{4e^{2t}}{9}$ (Choice C) C $\dfrac{2e^{2t}}{9}$ (Choice D) D $\dfrac{2e^{2t}}{3}$
Solution: We are asked to find the second derivative of a parametric function. Recall that the first derivative of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ is found with the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ Then, the second derivative is found with this following rule: $\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{\dfrac{d}{dt}\left(\dfrac{v'(t)}{u'(t)}\right)}{u'(t)}$ Let's start by finding $\dfrac{dy}{dx}$. $\dfrac{dy}{dx}=\dfrac{2e^{2t}}{3}$ Now we can find $\dfrac{d^2y}{dx^2}$. $\begin{aligned} \dfrac{d^2y}{dx^2}&=\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dx}{dt}\right)} \\\\\\\\\\ &=\dfrac{\dfrac{d}{dt}\left(\dfrac{2e^{2t}}{3}\right)}{\dfrac{d}{dt}(3t-6)} \\\\\\\\\\ &=\dfrac{\left(\dfrac{4e^{2t}}{3}\right)}{3} \\\\\\\\\\ &=\dfrac{4e^{2t}}{9} \end{aligned}$ In conclusion, $\dfrac{d^2y}{dx^2}=\dfrac{4e^{2t}}{9}$.